Solution
Step 1
Write the quadratic formula and use the formula to answer all the questions.
[tex]d\text{ = }\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
Step 2
Aubrey's
[tex]\begin{gathered} A_3(d)\text{ = }-0.25d^2+10d+3 \\ -0.25d^2+10d+3\text{ = 0} \\ Refine \\ -25d^2+1000d+300=0 \\ a\text{ = -25 , b = 1000 , c = 300} \\ d_{1,\:2}=\frac{-1000\pm\sqrt{1000^2-4\left(-25\right)\cdot\:300}}{2\left(-25\right)} \\ d_{1,\:2}=\frac{-1000\pm \:100\sqrt{103}}{2\left(-25\right)} \\ d_1=\frac{-1000+100\sqrt{103}}{2\left(-25\right)},\:d_2=\frac{-1000-100\sqrt{103}}{2\left(-25\right)} \\ d=-0.297,\text{ }\:d=40.29778 \\ d\text{ = 40.3ft} \end{gathered}[/tex]
Step 3
Kellen's
[tex]\begin{gathered} K_3(d)=-0.4d^2+13d+3.5 \\ Refine \\ -4d^2+130d+35=0 \\ d_{1,\:2}=\frac{-130\pm \sqrt{130^2-4\left(-4\right)\cdot \:35}}{2\left(-4\right)} \\ d_{1,\:2}=\frac{-130\pm \:6\sqrt{485}}{2\left(-4\right)} \\ d_1=\frac{-130+6\sqrt{485}}{2\left(-4\right)},\:d_2=\frac{-130-6\sqrt{485}}{2\left(-4\right)} \\ d=-\frac{-65+3\sqrt{485}}{4},\:d=\frac{65+3\sqrt{485}}{4} \\ d=-0.2670\:,\text{ }d=32.76703 \\ d\text{ = 32.8 ft} \end{gathered}[/tex]
Final answer
Aubrey's shot outside the platform because 40.3 > 37.
c) Kellen's is the closest
